Problem: $f(x, y) = \dfrac{x + \ln(x)}{\sin(y)}$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-\cos(y)}{\sin^2(y)} \left( x + \ln(x) \right)$ (Choice B) B $\dfrac{\left(1 + \dfrac{1}{x}\right)\sin(y) - \left( x + \ln(x) \right) \cos(y) }{\sin^2(y)}$ (Choice C) C $\dfrac{1 + \dfrac{1}{x}}{\sin(y)}$ (Choice D) D $\dfrac{-\cos(y)}{\sin^2(y)} \left(1 + \dfrac{1}{x} \right)$
Solution: Taking a partial derivative with respect to $x$ means treating $y$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial x} &= \dfrac{\partial}{\partial x} \left[ \dfrac{{x} + \ln({x})}{\sin(y)} \right] \\ \\ &= \dfrac{1}{\sin(y)} \dfrac{\partial}{\partial x} \left[ {x} + \ln({x}) \right] \\ \\ &= \dfrac{1 + \dfrac{1}{{x}}}{\sin(y)} \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial x} = \dfrac{1 + \dfrac{1}{x}}{\sin(y)}$